What is the sum from n=1 to infinity of 4/(n(n+4)) ?

The sum from n=1 to infinity of 4/(n(n+4)) is 25/12

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sum from n=1 to infinity of 4/(n(n+4))

n = 1 \sum \infty \frac{4}{n ( n + 4 )}

\frac{25}{12}

Solution steps

n = 1 \sum \infty \frac{4}{n ( n + 4 )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n ( n + 4 )}

Take the partial fraction of

\frac{1}{n ( n + 4 )} : \frac{1}{4 n} - \frac{1}{4 ( n + 4 )}

= 4 \cdot n = 1 \sum \infty \frac{1}{4 n} - \frac{1}{4 ( n + 4 )}

Expand telescoping series:

\frac{25}{48}

Simplify

4 \cdot \frac{25}{48} : \frac{25}{12}

= \frac{25}{12}

x \to 0 - lim (x^{3})

\frac{\partial}{\partial x} (\frac{3 y}{x ^{2} + y ^{2}})

\int (9 x (x^{8} - \frac{1}{9})) d x

\frac{d}{d x} (\frac{tan ( x ) - 4}{sec ( x )})

\frac{d}{d x} (\frac{7}{( x ^{2} - 1 ) ^{3}})

What is the sum from n=1 to infinity of 4/(n(n+4)) ?
The sum from n=1 to infinity of 4/(n(n+4)) is 25/12